Question

Prove that: sin^ 2 (45^ + theta)-sin^ 2 (45^ - theta)=2 sin theta.cos theta​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$sin^2(45^{\circ}+\theta)-sin^2(45^{\circ}-\theta)=2sin\theta cos\theta$

Formula Required :-

1)  sin²θ = (sinθ)²

2) sin(A + B) = sinAcosB + cosAsinB

3)  sin(A - B) = sinAcosB - cosAsinB

4) sin45° = 1/√2

5) sin²θ + cos²θ = 1

Solution :-

Taking L.H.S :-

$=sin^2(45^{\circ}+\theta)-sin^2(45^{\circ}-\theta)$

$=(sin(45^{\circ}+\theta))^2-(sin(45^{\circ}-\theta))^2$

[ ∴ sin²θ = (sinθ)² ]

$=(sin45^{\circ}cos\theta+cos45^{\circ}sin\theta)^2-(sin45^{\circ}cos\theta-cos45^{\circ}sin\theta)^2$

[ ∴ sin(A + B) = sinAcosB + cosAsinB

  sin(A - B) = sinAcosB - cosAsinB ]

$=\left(\left(\dfrac{1}{\sqrt{2}}\times cos\theta\right)+\left(\dfrac{1}{\sqrt{2}}\times sin\theta\right)\right)^2-\left(\left(\dfrac{1}{\sqrt{2}}\times cos\theta\right)-\left(\dfrac{1}{\sqrt{2}}\times sin\theta\right)\right)^2$

[ ∴ sin45° = 1/√2 ]

$=\left(\dfrac{cos\theta}{\sqrt{2}}+\dfrac{sin\theta}{\sqrt{2}}\right)^2-\left(\dfrac{cos\theta}{\sqrt{2}}-\dfrac{sin\theta}{\sqrt{2}}\right)^2$

$=\left(\dfrac{cos\theta+sin\theta}{\sqrt{2}}\right)^2-\left(\dfrac{cos\theta-sin\theta}{\sqrt{2}}\right)^2$

$=\dfrac{(cos\theta+sin\theta)^2}{(\sqrt{2})^2}-\dfrac{(cos\theta-sin\theta)^2}{(\sqrt{2})^2}$

$=\dfrac{cos^2\theta+sin^2\theta+2cos\theta sin\theta}{2}-\dfrac{cos^2\theta+sin^2\theta-2cos\theta sin\theta}{2}$

[ ∴ sin²θ + cos²θ = cos²θ + sin²θ = 1 ]

$=\dfrac{1+2sin\theta cos\theta}{2}-\dfrac{1-2sin\theta cos\theta}{2}$

$=\dfrac{1+2sin\theta cos\theta -(1-2sin\theta cos\theta)}{2}$

$=\dfrac{1+2sin\theta cos\theta - 1 + 2sin\theta cos\theta}{2}$

$=\dfrac{4sin\theta cos\theta}{2}$

= 2sinθcosθ

= R.H.S

Hence proved that 'sin²(45° + θ) -  sin²(45° - θ) = 2sinθcosθ '